LeetCode | 0451. Sort Characters By Frequency根据字符出现频率排序【Python】

LeetCode 0451. Sort Characters By Frequency根据字符出现频率排序【Medium】【Python】【桶排序】

Problem

LeetCode

Given a string, sort it in decreasing order based on the frequency of characters.

Example 1:

Input:
"tree"

Output:
"eert"

Explanation:
'e' appears twice while 'r' and 't' both appear once.
So 'e' must appear before both 'r' and 't'. Therefore "eetr" is also a valid answer.

Example 2:

Input:
"cccaaa"

Output:
"cccaaa"

Explanation:
Both 'c' and 'a' appear three times, so "aaaccc" is also a valid answer.
Note that "cacaca" is incorrect, as the same characters must be together.

Example 3:

Input:
"Aabb"

Output:
"bbAa"

Explanation:
"bbaA" is also a valid answer, but "Aabb" is incorrect.
Note that 'A' and 'a' are treated as two different characters.

问题

力扣

给定一个字符串，请将字符串里的字符按照出现的频率降序排列。

示例 1:

输入:
"tree"

输出:
"eert"

解释:
'e'出现两次，'r'和't'都只出现一次。
因此'e'必须出现在'r'和't'之前。此外，"eetr"也是一个有效的答案。

示例 2:

输入:
"cccaaa"

输出:
"cccaaa"

解释:
'c'和'a'都出现三次。此外，"aaaccc"也是有效的答案。
注意"cacaca"是不正确的，因为相同的字母必须放在一起。

示例 3:

输入:
"Aabb"

输出:
"bbAa"

解释:
此外，"bbaA"也是一个有效的答案，但"Aabb"是不正确的。
注意'A'和'a'被认为是两种不同的字符。

思路

桶排序

统计字符串字符出现的次数，然后桶排序，最后逆序输出字符串。

可以直接用 Python 的Counter类就能统计每个字符出现的次数，使用most_common函数就能按次序排列。

时间复杂度: O(n)
空间复杂度: O(m) m是桶的数量

Python代码

class Solution:
    def frequencySort(self, s):
        """ :type s: str :rtype: str """
        # # 法一：collections.Counter
        # import collections # 需要导入collections
        # count = collections.Counter(s).most_common() # Counter类就能统计每个字符出现的次数, 使用most_common函数就能按次序排列
        # res = ''
        # for char, cnt in count:
        # res += char*cnt # 字符与次数相乘
        # return res
        
        # 法二：桶排序
        import collections

        # 统计字符个数
        ret = []
        count_frequency = collections.defaultdict(int)
        for i in s:
            count_frequency[i] += 1  # 统计字符个数
        
        # 桶排序
        buckets = [[] for i in range(len(s) + 1)]
        for i in count_frequency:  # i是字符
            # print (i)
            buckets[count_frequency[i]].extend(i * count_frequency[i])
        
        # 打印排完序的字符串
        for i in buckets[::-1]:  # -1表示逆序
            if(i):
                ret.extend(i)
        return ''.join(ret)

        # return ''.join(c*-n for n,c in sorted((-s.count(c),c)for c in set(s))) # 一行代码也能解决

代码地址

GitHub链接